Answer:
[tex]\large\boxed{\dfrac{2-\frac{1}{y}}{3+\frac{1}{y}}=\dfrac{2y-1}{3y+1}}[/tex]
Step-by-step explanation:
[tex]\dfrac{2-\frac{1}{y}}{3+\frac{1}{y}}=\left(\dfrac{2y}{y}-\dfrac{1}{y}\right):\left(\dfrac{3y}{y}+\dfrac{1}{y}\right)=\dfrac{2y-1}{y}:\dfrac{3y+1}{y}=\dfrac{2y-1}{y}\cdot\dfrac{y}{3y+1}\\\\=\dfrac{2y-1}{3y+1}[/tex]
